Chapter 8 — Applications Of Integrals

Definition and geometric idea

The definite integral of a continuous function f(x) from a to b, written as ∫_a^b f(x) dx, represents the net signed area between the curve y = f(x) and the x-axis, over the interval [a, b]. Areas above the x-axis contribute positive value, areas below contribute negative value. The integral adds these signed contributions to give the net area.

Why this works

We approximate the area by summing areas of thin vertical rectangles of width Δx and height f(x_i). Letting Δx → 0 leads to the limit sum which defines the definite integral. The Fundamental Theorem of Calculus links this geometric area with antiderivatives: if F is an antiderivative of f, then ∫_a^b f(x) dx = F(b) − F(a).

Area between two curves

If f(x) ≥ g(x) for x in [a, b], the area of the region between the curves y = f(x) and y = g(x) is ∫_a^b (f(x) − g(x)) dx. This follows by subtracting the lower-surface signed area from the upper-surface signed area.

Net area vs total area

Net area = ∫_a^b f(x) dx. Total (geometric) area between curve and x-axis = ∫_a^b |f(x)| dx, which equals the sum of magnitudes of positive and negative parts.

Key geometric properties

• Additivity: ∫_a^b f(x) dx + ∫_b^c f(x) dx = ∫_a^c f(x) dx
• Reversal: ∫_b^a f(x) dx = −∫_a^b f(x) dx
• Linearity: ∫_a^b [αf(x) + βg(x)] dx = α∫_a^b f(x) dx + β∫_a^b g(x) dx

Interpretation in applications

Many real quantities are areas under a curve: displacement is area under a velocity-time graph, total accumulation (mass, charge, work) can be area under density or rate graphs, and probability (continuous) is area under a probability density function over an interval.

Signed area: The definite integral \(\int_a^b f(x)\,dx\) gives the signed (algebraic) area between the curve \(y=f(x)\) and the x-axis from \(x=a\) to \(x=b\). Areas above the x-axis count positive, areas below count negative. Thus the integral is the net (algebraic) area.

Actual (unsigned) area: The actual area (or unsigned area) between the curve and the x-axis is always nonnegative. It is obtained by integrating the absolute value of the function over the interval: \(A=\int_a^b |f(x)|\,dx\). In practice, you find the zeros (or sign changes) of \(f(x)\), split the interval at those points, and sum the absolute values of integrals on each subinterval.

Why they differ: If \(f(x)\) takes both positive and negative values on \([a,b]\), the signed integral may cancel positive and negative contributions and be smaller in magnitude (even zero), while the actual area sums magnitudes and is strictly nonnegative. Thus signed area measures net change; actual area measures total "amount" or size.

Procedure to compute actual (unsigned) area:

• Find points \(a = x_0 < x_1 < x_2 < \dots < x_n = b\) where \(f(x_i)=0\) or where sign of \(f\) changes.
• Compute integrals on each subinterval: \(I_i=\int_{x_{i-1}}^{x_i} f(x)\,dx\).
• Take absolute values and add: \(A=\sum_{i=1}^n |I_i| = \int_a^b |f(x)|\,dx\).

Area between two curves: The signed area of region between \(y=f(x)\) and \(y=g(x)\) is \(\int_a^b (f(x)-g(x))\,dx\). The actual area is \(\int_a^b |f(x)-g(x)|\,dx\). If one function lies above the other on the whole interval (say \(f(x)\ge g(x)\)), then the actual area simplifies to \(\int_a^b (f(x)-g(x))\,dx\).

Interpretations and cautions:

• Signed integral measures net quantity (e.g., displacement, net charge), while actual area measures total magnitude (e.g., distance traveled, total charge magnitude).
• A zero signed integral does not imply zero actual area.

Basic idea: The area under the graph of a continuous function y = f(x) between x = a and x = b is interpreted using the definite integral. If f(x) ≥ 0 on [a,b], the area of the region bounded by the curve, the x-axis and the vertical lines x = a and x = b is the definite integral ∫_a^b f(x) dx. The definite integral arises as the limit of Riemann sums (sum of rectangle areas) as the rectangle widths shrink to zero.

Signed vs geometric area: A definite integral gives a signed area: it is positive where the curve lies above the x-axis and negative where it lies below. To get the geometric (always positive) area you must take absolute value of the integrand or split the interval at the zeros of f.

When f changes sign: If f has zeros a = x_0 < x_1 < x_2 < ... < x_n = b, then the geometric area is

• Area = ∑_{k=1}^n ∫_{x_{k-1}}^{x_k} |f(x)| dx,
• or equivalently Area = ∑_{k=1}^n (±) ∫_{x_{k-1}}^{x_k} f(x) dx where the sign is chosen so each term is positive.

Area between two curves: If two functions satisfy f(x) ≥ g(x) on [a,b], the area of the region between them is ∫_a^b (f(x) − g(x)) dx. If they cross, find intersection points and split the interval.

Typical solution steps:

1. Sketch or identify where f(x) ≥ 0 and where f(x) < 0 on [a,b].
2. Find zeros/intersections to split the interval if needed.
3. Compute definite integrals on each subinterval.
4. Take absolute values or add positive contributions to get geometric area.

Short worked examples:

• f(x) = x^2 on [0,2]: area = ∫_0^2 x^2 dx = [x^3/3]_0^2 = 8/3.
• f(x) = x on [−1,1]: signed integral ∫_{−1}^1 x dx = 0, but geometric area = ∫_{−1}^0 (−x) dx + ∫_0^1 x dx = 1.
• f(x) = sin x on [0,π]: area = ∫_0^π sin x dx = [−cos x]_0^π = 2.

Connection to Riemann sums: Area = lim_{n→∞} Σ_{i=1}^n f(x_i^*) Δx where Δx = (b−a)/n and x_i^* is a sample point in the i-th subinterval. This limit equals the definite integral ∫_a^b f(x) dx when f is integrable.

Concept: The area between two curves y = f(x) and y = g(x) (on an interval [a,b]) is the area of the planar region bounded by the two graphs and the vertical lines x = a and x = b. If on [a,b] one function lies above the other (say f(x) ≥ g(x)), the area is the integral of the vertical distance between them:

Area = ∫_a^b [f(x) − g(x)] dx.

Derivation / Intuition: Using small vertical strips (width dx) the height of each strip is (top value − bottom value) = f(x) − g(x). Summing the areas of these strips and taking the limit gives the definite integral above.

Procedure (step-by-step):

• Find intersection points by solving f(x) = g(x). These give candidate limits a, b (order them so a < b).
• Determine which function is on top on each subinterval (graphical check or test point).
• If one function is always on top on [a,b], area = ∫_a^b [top − bottom] dx. If they cross, split the integral at crossing points and add absolute values (or use piecewise top minus bottom).
• As an alternative (when convenient), integrate with respect to y: if the region is better described by x = h1(y) and x = h2(y) with h2 ≥ h1 on [c,d], then area = ∫_c^d [h2(y) − h1(y)] dy.

Important notes:

• Always use the top function minus the bottom function so the integrand is nonnegative. If unsure, use |f(x) − g(x)| and split at zeros.
• If the region is bounded by more than two curves or requires piecewise description, split into appropriate subregions and sum areas.

What is it? When two plane curves are given as x = f(y) and x = g(y), the area of the region between them (for y between a and b) is found by integrating horizontal slices. Each slice at height y is approximately a rectangle of width |f(y) - g(y)| and infinitesimal height dy.

Derivation (idea): Consider a small horizontal strip at height y and thickness dy. Its length (along x) is the distance between the rightmost curve and the leftmost curve: x_right(y) - x_left(y). So the area dA of the strip ≈ (x_right(y) - x_left(y)) dy. Integrate from y = a to y = b (the intersection y-values) to get the exact area.

Formula: Area = ∫_a^b [x_right(y) − x_left(y)] dy = ∫_a^b [f(y) − g(y)] dy, where f(y) ≥ g(y) on [a,b]. If they cross inside the interval, split at crossing points and add integrals on subintervals.

Procedure (step-by-step):

• Express both curves as x = f(y) and x = g(y). If originally given as y = ..., solve for x in terms of y (if possible) or decide whether integrating w.r.t. y is advantageous.
• Find intersection points by solving f(y) = g(y). These give the limits a and b (or multiple subinterval endpoints).
• Determine which function is rightmost (larger x) and which is leftmost on each subinterval. Use test values or compare algebraically.
• Set up the integral(s): Area = ∑ ∫ (right − left) dy over each subinterval.
• Evaluate integrals and sum to get the total area.

Notes and tips:

• Always check if it is simpler to integrate with respect to y rather than x (e.g., when boundaries are naturally given as x = functions of y or when vertical slicing becomes awkward).
• If the region is closed by horizontal lines y = a and y = b and two x-curves, a single integral usually suffices.
• Remember absolute value: area is positive, so integrand must be (right − left) ≥ 0. If sign changes, split the integral at sign-change points.

Area of regions bounded by multiple curves is found by using definite integrals to add up infinitesimal strips that cover the region. The basic idea: choose an orientation of strips (vertical or horizontal) so that each strip's length is the difference between two curve-values, then integrate that difference between the appropriate limits (intersection points).

Step-by-step method

1. Sketch the curves and identify the closed region. Mark all intersection points (these give integration limits).
2. Decide whether vertical strips (integrate w.r.t. x) or horizontal strips (integrate w.r.t. y) are more convenient so that the top/bottom (or right/left) functions are single-valued on each subinterval.
3. On each subinterval (between consecutive intersections) determine which curve is above (or rightmost). If order changes, split the integral at the crossing.
4. Compute the area as the sum of integrals of (upper − lower) dx or (right − left) dy over the corresponding limits.
5. Check units and sketch the shaded area to verify correctness.

Important notes

• When more than two curves bound the region, partition the region into simpler subregions where just two curves bound each subregion.
• If it is easier, integrate w.r.t. y: use x = f(y) and x = g(y) and area = ∫(g(y) − f(y)) dy.
• Always use intersection points to get exact limits and split integrals wherever the ordering of curves changes.

When to use polar form

If the boundary curves are given naturally in polar form r = r(θ), use the polar area element: area = (1/2) ∫ r(θ)^2 dθ between the relevant angles. This is useful for petals, cardioids, circles centered at origin, etc.

What this topic is about: When you compute areas or other integrals for regions bounded by curves, you must first determine the correct limits (bounds) of integration and draw a clear sketch of the region. Limits come from the intersection points of the bounding curves or from the given domain along an axis.

Step-by-step method

• 1. Draw a rough sketch of all curves involved (plot key points, intercepts, and shape). A sketch helps decide which curve is above/below or left/right.
• 2. Find intersections by solving the equations of pairs of curves. The real x (or y) coordinates of intersection give candidate limits.
• 3. Decide the direction of integration (dx or dy): use vertical strips (integrate w.r.t x) when region is best described as y between two functions y = f(x) and y = g(x); use horizontal strips (integrate w.r.t y) when region is best described as x between two functions x = h(y) and x = k(y).
• 4. Identify upper/lower or right/left curves on each subinterval between intersection points (test a sample x or y value to see which function is larger).
• 5. Split the integral if necessary when the upper/lower relationship changes inside the interval. Each subregion gets its own integral.
• 6. Set up integrals using the appropriate bounds and integrand (typically difference of outer and inner functions).

Common checks and tips

• Check symmetry: if the region is symmetric about the y-axis (or x-axis), you can integrate from 0 to a and double the result.
• If curves are given implicitly (like circles), solve for y (or x) to express the boundary used for the chosen direction of integration.
• Remember the sign: area is positive; use absolute value or ensure you always subtract lower from upper (or left from right).

How a sketch helps: A clear graph annotated with intersection points, shaded region, and a representative strip (vertical or horizontal) prevents sign and bound errors and shows whether you must split the integral.

Idea: When two curves cross, the one on top (larger y) can change within the domain. Area between them from x = a to x = b equals the integral of the absolute difference: area = ∫_a^b |f(x) - g(x)| dx. To evaluate this, find the x-values where f(x)=g(x) (intersection points), split [a,b] into subintervals at those points, and on each subinterval integrate (top − bottom).

Step-by-step method

1. Find intersection points: solve f(x) = g(x). Let the real solutions inside [a,b] be x₀=a < x₁ < x₂ < … < x_n = b.
2. On each subinterval (x_i, x_i+1), determine which function is on top. You can test a sample x in the interval and compare f(x) and g(x).
3. Compute area as sum of integrals where integrand is (top − bottom):
       Area = Σ_i=0^n-1 ∫_{x_i}^x_{i+1} (f_top(x) − f_bottom(x)) dx
   Equivalently Area = ∫_a^b |f(x) − g(x)| dx.
4. If functions are given as x = h(y), do the same with integrals in y: Area = ∫ (right − left) dy, splitting at y-values of intersection.

Common pitfalls

• Forgetting to split at each intersection: this may give a signed integral of zero instead of total area.
• Using the wrong top − bottom sign on a subinterval: always check by sampling a point.
• Neglecting multiple intersections (more than two): split at all intersection x-values inside [a,b].

Why absolute value works: The integrand f(x) − g(x) can be positive or negative. The definite integral ∫ (f−g) dx gives a signed net area; taking |f−g| and splitting into intervals makes every piece positive, which is the geometric area.

When to use splitting: Always when f and g cross inside the integration limits (i.e., when f(x) − g(x) changes sign). Splitting makes each integrand nonnegative.

What is symmetry in this context? In area calculations using integrals, symmetry means the geometric or algebraic invariance of a curve/region under a reflection or rotation. Common useful symmetries are about the y‑axis (even functions), about the origin (odd functions), and about a vertical line x = c. Recognizing symmetry reduces the interval of integration or the number of pieces you must compute.

Key ideas:

• Even function: f(−x) = f(x). A region under an even function on [−a, a] is symmetric about the y‑axis; the total area is twice the area on [0, a].
• Odd function: f(−x) = −f(x). The signed integral ∫_{−a}^{a} f(x) dx = 0. For areas (nonnegative quantities) you must treat absolute values, but signed cancellation can simplify net area calculations.
• Symmetry about a vertical line x = c: if f(2c − x) = f(x), the curve is symmetric about x = c. Then area over [c − a, c + a] equals twice the area over [c, c + a] (or over [c − a, c] depending on which half you compute).
• Polar symmetry: For polar curves r = r(θ), symmetry relations like r(−θ) = r(θ) (symmetry about the initial line) or r(π − θ) = r(θ) (symmetry about the vertical line through the pole) let you compute area of one symmetric sector/petal and multiply.

How to use symmetry in area integrals:

1. Identify the axis/point/line of symmetry by checking algebraic relations (even/odd or shifted symmetry) or by inspection of the curve.
2. Reduce the integral domain to one symmetric piece (usually from 0 to a or c to c + a).
3. Compute the integral over that smaller interval and multiply by the symmetry factor (usually 2, or the number of symmetric sectors/petals).

Advantages: fewer integrals, simpler limits, sometimes cancellation of odd terms making integrals zero, and easier evaluation of complicated areas (e.g., petals of a rose, ellipse halves, circle halves).

Notes on signed vs geometric area: The definite integral ∫ f gives signed area. When using symmetry for geometric area, take absolute values where needed: geometric area = ∫ |f(x)| dx or split into regions where f keeps sign.

Overview: In area problems we use definite integrals to compute the area under a curve, between two curves, or between a curve and the axis. Several algebraic properties of definite integrals simplify these computations and help set up the correct integrals.

Key properties and how they apply to area:

• Linearity: ∫_a^b [αf(x) + βg(x)] dx = α∫_a^b f(x) dx + β∫_a^b g(x) dx. This lets you split combined shapes into simpler parts (e.g., separate polynomial pieces) and scale areas by constants.
• Additivity (interval splitting): ∫_a^b f(x) dx = ∫_a^c f(x) dx + ∫_c^b f(x) dx for any c between a and b. Use this to compute composite areas by splitting at intersection points or axis crossings.
• Reversal of limits: ∫_a^b f(x) dx = −∫_b^a f(x) dx. Useful when changing integration order or matching orientation of intervals.
• Sign property: If f(x) ≥ 0 on [a,b], then ∫_a^b f(x) dx ≥ 0. If f(x) ≤ 0, the integral ≤ 0. For area you must use the absolute value where the curve goes below the axis (area is non‑negative).
• Area with sign vs. geometric area: The definite integral gives a signed area. The geometric area between the curve y = f(x) and the x-axis on [a,b] is Area = ∫_a^b |f(x)| dx. If f does not change sign on [a,b], Area = |∫_a^b f(x) dx| = ∫_a^b f(x) dx (if f ≥ 0).
• Area between two curves: If f(x) ≥ g(x) on [a,b], Area = ∫_a^b [f(x) − g(x)] dx. Use intersection points to find limits a and b.
• Symmetry:
  • If f is even (f(−x)=f(x)), then ∫_−a^a f(x) dx = 2∫_0^a f(x) dx. This halves the work for symmetric regions about the y-axis.
  • If f is odd (f(−x)=−f(x)), then ∫_−a^a f(x) dx = 0 (signed area cancels). For geometric area use ∫_−a^a |f(x)| dx = 2∫_0^a |f(x)| dx.
• Substitution (translation): ∫_{a+h}^{b+h} f(x) dx = ∫_a^b f(t+h) dt (or by change of variable). This helps shift intervals without changing shape of area.
• Comparison/Monotonicity: If f(x) ≥ g(x) on [a,b], then ∫_a^b f(x) dx ≥ ∫_a^b g(x) dx. Useful to bound areas.

How to use these in area problems — procedure:

1. Draw the curves and find intersection points to determine limits.
2. Decide which function is on top (f ≥ g) on each subinterval; split the interval if order changes.
3. Use Area = ∫_a^b (top − bottom) dx. If the curve crosses the x-axis, compute area as the sum of integrals of |f| on each sign-constant subinterval.
4. Use linearity, symmetry, and additivity to simplify calculations.

Note: For areas using horizontal slices integrate with respect to y: Area = ∫_c^d [x_right(y) − x_left(y)] dy.

Overview: In Applications of Integrals you convert a geometric/physical quantity (area, volume, length, mass, work, centroid, etc.) into a definite integral by modelling the quantity as a limit of sums of small contributions (slices, washers, shells, strips). The core idea is the “small-slice” method: approximate the object by many small pieces whose contribution you can compute, express each piece as a function of a variable, sum (integrate) and take limits.

General problem-solving strategy (step-by-step):

• 1. Read the problem carefully and identify what quantity is required (area, volume, length, work, mass, centroid, etc.).
• 2. Sketch the region/solid and label axes; mark boundaries and intersection points.
• 3. Choose the slicing method: slices perpendicular to x (use dx) or y (use dy); or choose method of disks/washers or cylindrical shells for volumes.
• 4. Express the small element (dA, dV, dL, dM, dW, dS) in terms of the variable: geometry of slice (height, radius, thickness, density).
• 5. Determine correct limits (intersection points or given bounds). Use symmetry to halve/double the work if applicable.
• 6. Set up the definite integral; simplify integrand algebraically (substitution if helpful).
• 7. Evaluate the integral; interpret units and check result for reasonableness (dimensions, limiting cases, sign).
• 8. State the final answer clearly, include units.

Tips & common pitfalls:

• Decide early whether dx or dy gives simpler bounds/expressions.
• Use symmetry (even/odd, axis symmetry) to simplify integrals or reduce limits.
• Watch orientation: when integrating (upper − lower) ensure functions are correctly identified on the interval.
• Keep units consistent (length, area, density units) when computing physical quantities (mass, work).
• Sketch a representative slice and label its dimensions — that avoids many mistakes.

Common integral-model types (how to recognize):

• Area between curves: integrate (top − bottom) dx (or right − left) dy.
• Volume of revolution: disk/washer method (cross-sections perpendicular to axis) or shell method (cylindrical shells).
• Volume with known cross-sections: V = ∫ A(x) dx where A(x) is area of cross-section.
• Arc length: integrate sqrt(1 + (dy/dx)^2) dx (or parametric form).
• Surface area of revolution: 2π ∫ radius * arc-element = 2π ∫ y sqrt(1 + y'^2) dx (or corresponding dy form).
• Mass / center of mass: integrate density × length/area element; centroid via moments: x̄ = M_y / A, ȳ = M_x / A.
• Work / hydrostatic forces: integrate force of each slice (pressure×area×distance) across slices.
• Average value / probability density: (1/(b−a)) ∫ f(x) dx; probability: ∫ p(x) dx = 1 over domain.

How to present solutions (recommended): include a clear sketch, write the element expression (dA, dV, dL...), state limits, show integral set-up, evaluate and give units and brief interpretation.